Question #588d8

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Question #588d8

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Answer 1 · 2017-05-22T21:32:39

Use the identities:

$sin (x + y) = sin (x) cos (y) + cos (x) sin (y)$ $sin(x+y)=sin(x)cos(y)+cos(x)sin(y)$
$cos (x - y) = cos (x) cos (y) + sin (x) sin (y)$ $cos(x-y)=cos(x)cos(y)+sin(x)sin(y)$
${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

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We will start with the LHS:

$sin (x + y) cos (x - y)$ $sin(x+y)cos(x-y)$

$(sin x cos y + cos x sin y) (cos x cos y + sin x sin y)$ $(sinxcosy+cosxsiny)(cosxcosy+sinxsiny)$

Now use the FOIL method (from the good ol' days back in Algebra 1)

$sin x cos x {cos}^{2} y + {sin}^{2} x sin y cos y + {cos}^{2} x sin y cos y + sin x cos x {sin}^{2} y$ $sinxcosxcos^2y+sin^2xsinycosy+cos^2xsinycosy+sinxcosxsin^2y$

$({sin}^{2} y + {cos}^{2} y) (sin x cos x) + ({sin}^{2} x + {cos}^{2} x) (sin y cos y)$ $(sin^2y+cos^2y)(sinxcosx)+(sin^2x+cos^2x)(sinycosy)$

Now use the third identity listed above.

$sin x cos x + sin y cos y$ $sinxcosx+sinycosy$

In fact, $sin (x + y) cos (x - y)$ $sin(x+y)cos(x-y)$ does NOT always equal ${sin}^{2} x - {sin}^{2} y$ $sin^2x-sin^2y$ . It DOES however equal $\frac{1}{2} (sin 2 x + sin 2 y)$ $1/2(sin2x+sin2y)$ , if that was somehow what was meant to be written.

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Question #588d8

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